Question

$\text{Prove that}\frac{\cos 7x+\cos 5x}{\sin 7x-\sin 5x}=\cot x$

Answer 1

L.H.S=$\frac{\cos 7x+\cos 5x}{\sin 7x-\sin 5x}$

$\cos A-\cos B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$

$\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$

$=\frac{2\cos \frac{7x+5x}{2}\cos \frac{7x-5x}{2}}{2\cos \frac{7x+5x}{2}\sin \frac{7x-5x}{2}}$

$=\frac{2\cos \frac{12x}{2}\cos \frac{2x}{2}}{2\cos \frac{12x}{2}\sin \frac{2x}{2}}=\frac{2\cos 6x\cos x}{2\cos 6x\sin x}$

$=\frac{\cos x}{\sin x}=\cot x$

Hence proved